$AB=5,0;\; AC=3,0\;$and $BC=4,0$. What is the height?
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$\begingroup$Compute the area of triangle in two ways. One way is $\frac{1}{2}(h)(5)$ and the other way is $\frac{1}{2}(3)(4)$.
$\endgroup$ $\begingroup$Let $D$ be the point where $h$ strikes $AB$ .Then let $AD=y\implies DB=5-y$
Now $(5-y)^2+h^2=16;y^2+h^2=9$
Subtracting we get $9-10y=-9\implies y=\dfrac{9}{5}$
Hence $AD=\dfrac{9}{5}\implies h=\dfrac{12}{5}$
$\endgroup$ $\begingroup$METHOD $1$.
Suppose the perpendicular drops on a point $P$ (on line $AB$). Then suppose the length $AP$ is $x$ and $BP$ ultimately becomes $5-x$ .
Now use Pythagoras theorem and you will have $4^2-h^2=(5-x)^2$ and $3^2-h^2=x^2$. You have two equations and two variables.
From here you can go further as: $4^2-h^2-(3^2-h^2)=(5-x)^2-(x^2)$ which gives $x=9/5$.
Since $3^2-h^2=x^2$ thus $h= 12/5$.
METHOD $2$.
As known area of triangle is $\frac{(ab)sinC}{2}$ and is also equal to $\frac{hc}{2}$.
THus $\frac{(ab)sinC}{2}$=$\frac{hc}{2}$. Putting $a=4,b=3,c=5$ and $sinC=1$ we get that $h=12/5$.
METHOD $3$.
Since you know all the sides, you can use Heron's formula to get area as $AREA=\sqrt{s(s-a)(s-b)(s-c)}$. Then equate it to $\frac{hc}{2}$ and you will get $h=12/5$ again.
$\endgroup$ 1 $\begingroup$You can find the area of the triangle by using Hero's formula $\sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac{a+b+c}{2}$ and then substituting area and base in $Area=\frac{1}{2}*base*height$
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